1. Hamiltonian formalism
To formulate Hamiltonian mechanics at a minimum we need a Poisson bracket on the smooth functions of a manifold. Poisson manifolds, as they are called (see [1]), come with a (skew-symmetric) bivector field \(\pi\); its rank as a map \(T^*X \to TX\) stratifies \(TX\) on the even dimensions \(0, 2, \dots, 2n\) (due to the bivector field being skew-symmetric.) Each stratum folliated by symplectic leaves. It is therefore natural (and simpler) to begin with symplectic manifolds when studying the Hamiltonian formalism.
Each function \(H\in C^\infty(X)\) defines the evolution of every other smooth quantity, say \(f\in C^\infty(X)\), by the evolution equation
\begin{align} \dot{f} & = \{H, f\}. \end{align}This equation means that \(H\) defines1 curves \((x(t), \dot{x}(t))\) with any given initial data such that \(\frac{d}{dt}f(x(t), \dot{x}(t)) = \{H, f\}(x(t), \dot{x}(t))\). These curves are defined by the same evolution equation, yielding Hamilton's equations:
\begin{align} \frac{d x}{dt} = \frac{\partial H}{\partial \dot{x}}, \quad \frac{d \dot{x}}{dt} = -\frac{\partial H}{\partial x}. \end{align}The function \(H\) is the energy of the system. We can play the same "game" for any other function \(g\in C^\infty(X)\) even if it is not an energy function; having fixed \(H\) (and hence our system), if we then look at the \(g\)-evolution, we will (most likely) traverse through the states of different phase-space trajectories. This can be useful because the algebraic behavior of \(g\) with respect to the bracket can give us information on the geometric features of such a traverse path without solving any differential equations!
If a function \(f\) satisfies \(\{f, H\} = 0\) it is said to be a first integral of the Hamiltonian system. It is useful to find as many functions in involution with \(H\) as one can. In a complicated problem, such as the Kepler problem, one simply writes down the angular momentum vector \(\boldsymbol{L}\) and the (normalized) Runge-Lenz vector \(\boldsymbol{A}\) and it turns out that the brackets of their coordinate functions close over themselves, which has immediate implications for the geometry of the problem.
1.1. TODO add example here
This example should show the \(\boldsymbol{L}\cdot\boldsymbol{\delta\omega}\), etc, examples from Maclay.
1.2. TODO remove dot-x notation
Remove this notation because in a general symplectic manifold \((X, \omega)\) the notation is bad (gets conflated with \(TX\)) instead one should use \((z^1, \dots, z^{2n})\) as is customary.
2. Lagrangian formalism
2.1. Classical treatment
Let \(X\) be an \(n\)-dimensional manifold and \(TX\) its tangent bundle with local coordinates \((x^\lambda, \dot{x}^\mu)\) and let \(L(t, x^\lambda, \dot{x}^\mu)\) be a (first order) Lagrangian density. The action integral is given by \(S \coloneqq \int_{t_0}^{t_1} L dt\).
Given a path \(x \colon I \to X\) for the interval \(I \coloneqq [t_0, t_1]\), lying inside the coordinate patch, we can form a variation of it as follows: consider any path \(\eta \colon (I, \{t_0, t_1\}) \to (\mathbb{R}^n, 0)\). There's a family of paths \(r \mapsto (x^\lambda)^{-1} \circ (x(t) + r\eta(t))\) which induces the Gateaux derivative at \(\left.\frac{d}{dr}\right|_{r=0}\) denoted by \(\delta x / \delta \eta \colon (I, \{t_0, t_1\}) \to (TX, 0)\). Due to the fact that we will consider every such \(\eta\), we may suppress it from notation and finally denote the variation of \(x\) as \(\delta x\), which then denotes any path \((I, \{t_0, t_1\}) \to (TX, 0)\). Since the action \(S\) is a function of the path \(x(t)\), it too has a variation \(\delta S\), which we may compute as:
\begin{align} \delta S = \int_{t_0}^{t_1} (\partial_\lambda L \cdot \delta x^\lambda + \partial_\mu L \cdot \delta \dot{x}^\mu) dt. \end{align}We may use integration by parts on the second term to obtain:2
\begin{align} \delta S = \cancel{\left.\partial_\mu L\cdot \delta x^\mu\right|_{t_0}^{t_1}} + \int_{t_0}^{t_1} (\partial_\lambda L - \frac{d}{dt}\dot{\partial_\lambda} L) \delta x^{\lambda} dt. \end{align}Setting \(\delta S = 0\) for all variations \(\delta x\) yields exactly the Euler-Lagrange equation:
\begin{align} \partial_\lambda L - \frac{d}{dt}\dot{\partial}_\lambda L = 0. \end{align}Note that \(\frac{d}{dt} \dot{\partial}_\lambda L = \partial_t \dot{\partial}_\lambda L + \dot{x}^\kappa \partial_\kappa \dot{\partial}_\lambda L + \ddot{x}^\mu \dot{\partial}_\mu \dot{\partial}_\lambda L\). We will use this expression later to define the total derivative when we discuss Lagrangians in the Jet bundle formalism.
2.2. Jet bundle formalism
In the above, the path \(\delta \dot{x}\) is a section of \(x^*(VTX)\). As we can see, we arrive at the second order jet bundle in this manner. Replacing the \(x\)-coordinate with a \(y\)-coordinate and generalizing the time coordinate to any base \(x\)-coordinate, we arrive at the formalism on \(J^1 Y \to X\), with a first-order Lagrangian density \(L(x^\lambda, y^i, y^i_\lambda)\omega\) for some top-level differential form \(\omega\) on \(X\).
The Euler-Lagrange operator is obtained by the same means, that is via variations \(\delta y\) and setting \(\delta S = 0\), to be
\begin{align} \mathcal{E}_i(L) \coloneqq \partial_i L - d_\lambda \partial^\lambda_i L. \end{align}Continue with this chatgpt conversation. See also https://chatgpt.com/s/t_6a1ff62723ec819195876a2ff66418a0.
3. TODO Things to study
- ☐ See also Ian Anderson's "The Variational Bicomplex" and his youtube series https://www.youtube.com/watch?v=oUagwAW3wUQ.